Question

Tính P = 1²-2²+3²-4²+...+99²-100²

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Answer 1

\(P=1-2^2+3^2-4^2+...+99^2-100^2\)

\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(99-100\right)\left(99+100\right)\)

\(=-\left(3+7+...+199\right)\)

P có số phần tử  \(\frac{199-3}{4}+1=50\)

\(-P=\frac{50\left(199+3\right)}{2}=5050\)

\(\Rightarrow P=-5050\)

 

Answer 2

Đặt A=1²-2²+.....-2016²

=> -A=2²-1²+4²-3²+6²-5²...+2016²-2015²

-A=(2-1)(2+1)+(4-3)(4+3)+(6-5)(6+5)...+(2016-2015)(2016+2015)

-A=3+7+11+...+4031

-A=[(4031-3):4+1]:2 x (3+4031)

-A=2033136

A=-2033136